It feels to me like there might be some confusion between different surfaces here. I will not try to show tricks of how to prove bound currents do not lead to charge accumulation, and what happens with them on the surface. Instead, I will introduce surface current through the route of continuity conditions, and then show what is necessary for it to not lead to charge accumulation. You can define it as bound surface current if you wish.
For things like charge density $\rho$ and current density $\mathbf{J}$ you can establish existence of conservation laws of the shape:
$$ \boldsymbol{\nabla}.\mathbf{J}+\partial_t \rho=0 $$
Where $t$ is time. You can integrate the above equation over some volume $V$, with surface $\partial V$ to find:
$$ \int_V d^3 r\,\partial_t \rho=\frac{d}{dt} \int_V d^3 r\, \rho=\frac{dQ_V}{dt}=-\int_V d^3 r\, \boldsymbol{\nabla}.\mathbf{J}=-\oint_{\partial V} d^2 r\, \mathbf{\hat{n}}.\mathbf{J} $$
Where $Q_V$ is the charge in volume $V$, and $\mathbf{\hat{n}}$ is the normal to the surface pointing from inside to outside. Note that $\mathbf{J}$ here is still a fully-fledged 3d surface current density with units of Amps/meter-squared.
When it comes to surface current density, one is usually concerned with continuity of Maxwell's equations across the domains. In particular, we have:
$$ \boldsymbol{\nabla}\times\mathbf{H}=\mathbf{J}+\partial_t \mathbf{D} $$
For electric displacement $\mathbf{D}$ and magnetizing field $\mathbf{H}$. This is the only equation to contain current density $\mathbf{J}$.
Consider integrating this equation, both sides of it, over a surface of a rectangle $R$ that straddles two domains. Choose coordinates in such a way that domain boundary is XY plane, with $z>0$ corresponding to domain (1) and $z<0$ corresponding to domain (2). Let the $R$'s center be positioned at the origin. Let it lie in the $XZ$ plane. The size along the $X$ dimension, call it length, is $l$, the size along the $Z$ dimension, call it height, is $h$
We will assume that $h$ and $l$ are small enough for domain boundary to be nearly parallel to XY plane in the region concerned. I don't think there are any interesting effects due to any remaining curvature.
Choose, $l\gg h$, then:
$$ \int_{R} d^2 r\,\mathbf{\hat{y}}.\boldsymbol{\nabla}\times\mathbf{H}=l\cdot\left(\mathbf{H}^{(1)}-\mathbf{H}^{(2)}\right)\Big|_{origin}.\mathbf{\hat{x}}+\mathcal{O}\left(h\right)=\int_{R} d^2 r\,\mathbf{\hat{y}}.\mathbf{J}+\mathcal{O}\left(h\right) $$
The term with electric displacement vanishes unless we assume $\mathbf{D}$ to be infinite or change infinitely fast. So we end up with discontinuity in magnetization field to be dependent on current density integral not vanishing even in the limit $h\to 0$. One way to do this is to define current density as:
$$ \mathbf{J}=\mathbf{J}^{(vol)}+\mathbf{J}^{(s)} $$
With former vanishing inside the integrals of the kind above, and the latter taking the form $\mathbf{J}^\left(s\right)\to \delta\left(z\right)\mathbf{K}\left(x,y\right)$ near the location we have considered. Here $\mathbf{K}$ would have units of Amps/meter, and would be defined only on the domain boundary. It would probably also make sense to have its direction to be constrained to pointing only along the domain boundary.
If you wanted the surface current density to be associated with no charge accumulation, you would have:
$$ \boldsymbol{\nabla}.\mathbf{J}^{(s)}\to \delta\left(z\right).\left(\partial_x K^{(x)}+\partial_y K^{(y)}\right)=0 $$
So as long as you $\mathbf{K}$ lies inside the domain interface and satisfies 2d version of the divergence condition, you will have no charge density accumulation.
ADDENDUM
In case anyone wants a more generic treatment, I think it goes like this. Let $f\left(\{x^{(i)}\}\right)=0$ be the condition that describes the interface ($\{x^{(i)}\}=x,y,z$).
A suitable expression for current density two-form is then (see exterior derivative):
$$ \mathbf{J}=K_a\left(\{x^{(i)}\}\right) \cdot \delta\left(f\left(\{x^{(i)}\}\right)\right)\:dx^{(a)}\wedge df $$
Which guarantees that surface current density ($K_a$) one-form does not have a component perpendicular to the surface. The zero-divergence condition then becomes:
$$ d\mathbf{J}=\partial_a K_b\,\delta\left(f\left(\{x^{(i)}\}\right)\right)\:dx^{(a)}\wedge dx^{(b)}\wedge df=0 $$
Which is fulfilled if $\epsilon^{abc}\,\partial_a K_b\,\partial_c f=0$ (using Levi-Civita relative tensors).